 So, quickly, very simple problem, so the idea is that the way for example, there is this one thing which is a big rage now, that our students to learn from their neighbors predominantly like especially during tutorials when they are allowed to interact mainly because whatever doubts they have are the doubts which we may not be able to anticipate, whereas like they can share each other concerns in a much better way. With that aside, what we will quickly go through is what do we emphasize the students that the way we are going to solve this problem. First we draw the free body diagram, clearly as we had discussed earlier, we see that this is a fixed support. Now, one of the errors that many students do is they forget that this is a fixed support and they will take that as a hinge and because of that what happens is that they will have only E x and E y and they will not have this moment which is necessary for this fixed support. Now, what happens is that we are specified what is the tension in the cable. So, if the tension in the cable is not specified, then the current problem the way it is given has three unknowns here and one extra unknown plus 3 plus 1, 4 unknowns. So, this becomes a statically indeterminate problem, but we are given what the tension in the cable is. So, with these three reactions, this becomes a statically determinate problem. On the other hand, we take this cable tension to be 150 kilo Newton and keep this as a hinge, then we have only two unknowns and three equations and you will see that a system will not be a stable system. There will be multiple solutions for the system which in other words mean that a system is not in equilibrium. So, this is the most important point that the free body diagram has this moment and once we have that, one way to do is for example, we emphasize the students that what are the mistakes that they can make. So, this is, if you write an equation like this, sigma fx equal to 0, if you say that this is Ex plus cos 36.9 into 150 kilo Newton, so which is this angle, so this is not taking an appropriate equation for equilibrium, whereas what we do is that that we know that all these distances are given to us, this is 4.5, 6 and the hypotenuse will be 7.5, so 4.5 by 7.5 is the sign of this angle which will be the horizontal component and if you do 6 by 7.5, so one of the mistakes students make is look that this angle is 150 kilo Newton, so strictly speaking for equilibrium in the horizontal direction, we want 4.5 divided by 7, but many times students say it is 6 divided by 7.5, so what we have to emphasize, so one small trick to emphasize here is that that if this cable were perfectly straight, which means that the vertical distance is very small and horizontal distance is very large, then essentially this value becomes 1. In the other extreme, when the vertical distance is very large compared to the horizontal distance, then this value becomes 0, so essentially what matters for finding out a component in the x direction for this cable is this distance 4.5, so this is very common mistake the students will write 6 divided by 7.5 and that is not appropriate and we have to again see that the signs that we have chosen that this is ex which is plus, so if we take the component of fx that will also has the same direction as ex, so that also should be positive whereas many times students make mistake with this the plus will they will make as negative, so those are the points that we want to emphasize and when we emphasize that all these equations are perfectly fine, this cos why was it wrong because if we take this theta 4.5 by 7.5 will be sin inverse of 36.5, so if you do cos then they are taking the vertical components, so these are the common mistakes that students make which you want to make sure, we want to emphasize that these are the common mistakes that you students can make and we want to make sure that they know what is happening and again as I had discussed earlier that we tell them that what is the negative sign signify, it just means that we have to repeatedly tell them that ex direction we assume it to the right but it is really to the left, is that what we have to emphasize. Now the idea is that if the students has taken this as the hint support then they are gone because this Me is not there but truly speaking for this system to be in equilibrium we have to first now if you want to find out sorry if you want to find out what is the equilibrium in the y direction okay, so these are the two appropriate choices and the same mistakes which students had done earlier that the vertical, vertical clearly now what are the components we are taking f y, y as positive, so e y is positive this is fine okay but if you take minus 4 into 2, 20 kilo Newton that is also great okay but if you do plus 6 by 7.5, 6 by 7.5 into 150 is perfectly right the vertical component but then that should be minus because it is pointing downwards, so students make these mistakes quite often, so with all these things in mind okay, so this will be the right equation e y in the vertical direction minus 4 into 20 in the downward direction so minus, minus cos 36.9 this angle is 36.9 you can verify on the calculator okay or it is equal to 6 divided by 7.5 you can write and that into 150 is fine or if you write 6 by 7.5 with a minus sign because this is acting downwards and our y is taken to be positive then this is the right equation of equilibrium okay again the same thing about positive and negative sign okay and another thing that we make them discuss with the neighbors so that they know okay what mistakes they are making because very likely the neighbor and the student would be likely to face the same problem so it is a good idea but again that is a optional thing the main thing is to emphasize them that what are the appropriate conventions that they should be choosing now ultimately we want to take torque about e and when we take torque or moment about e whatever okay torque moment are equivalent things we can use whatever we like then the components are what so it is this 20 into 18 into 1.8 into 4 that is 7.2 how will it act it will act in the anticlockwise direction plus 20 into 5.4 plus 20 into 3.6 plus for the last force 20 into 1.8 okay so these are all the torques of this 20 kilo Newton forces about this and then finally what do we want now this is one very interesting point that this force can be resolved into two components one is a vertical component one is a horizontal component now the horizontal component mind well okay horizontal component passes through e so that will have no contribution to moment about e whereas the vertical force has a moment arm of 4.5 and what direction the torque will be will be in the clockwise direction and that is why we take a negative sign here because everything because the sign convention that we have chosen is anti anticlockwise is positive so that is a negative sign now note one thing somebody has asked previously the principle of transmissibility now this 4 now instead of breaking it here you can say that the 4 so this tension 150 kilo Newton force is acting at point d and now when you break it there you will have a horizontal component and a vertical component now the vertical component passes through point e and that will not has any that will not have any contribution so the only contribution will be now this horizontal force into 6 and you substitute all the values you will see that this 4.5 will come here and this 6 will go there so whatever you get the moment about point e you can either resolve the force here or resolve the force here you will get the same answer so that is one another point which is like another demonstration of principle of transmissibility that you can emphasize to the students okay so this is a brief discussion okay if you want to add or point out anything you can have a two minutes brief discussion and then move on to the next problem 1 1 3 9 if constraints are more more than 2 3 4 5 6 like that and if those are passing through single point yes so that is that is the point so for example you totally right okay it's a very good point that you are making yes so I can do a very simple example like this so for example I have rollers like this okay coming like this okay then I have another roller which is acting like this okay these are all perpendicular then I have a third roller okay which is acting like this and I may have more more of them this and I may have huge number of such constraints okay like this and that is what I think you have in mind and if I draw the free body diagram I am not removing the roller right now for just saving time the reactions will look like this like this like this like this like this like this okay so we have five reactions so nively if we think that five reactions it's a statically indeterminate structure then we are wrong as you rightly pointed out so having five reactions doesn't mean that your structure is stable but as we had pointed out briefly that even if I five reactions they all meet at a point now to this structure if I apply a force like this then the structure will be unstable why because there is nothing to balance the moment that this force has about this point oh so as I rightly pointed out just having more constraints doesn't mean that the structure is stable but what happens is that if you have more constraints and the structure is properly constrained then it becomes statically indeterminate like for example in our problem we have this and this is 150 but in this problem additionally now this is a stable system clearly okay this is a very stable system but additionally if I also put a roller like this then the number of unknowns becomes 3 here plus 1 reaction here and so this is a 1 degree of statically indeterminate structure okay it has 1 degree of indeterminacy okay I think that answers your question hopefully okay so we can move on now okay we'll move on to the next problem now there is one thing I just wanted to emphasize is that that when we did moment equilibrium we could have taken chosen any point you want to we could have chosen a b f or some any point in Timbuktu but the reason we chose e for the simple fact is that that these two unknown forces to begin with which were unknown to begin with but which we found out by calculation they don't contribute to the moment about point e for this free body diagram okay now what you can do is that there can be a check that you can use that if you gotten some answers are those answer right or not what we can do we can use these values and then take talk about say point f for the entire set of forces acting on this free body diagram and any for answers are right then that talks also come out to be 0 so that is a consistency check that we can use okay so we'll now move on to the next problem let us move to the third one which is a little bit simpler okay and this problem is there for a reason I want to emphasize one point it's a very simple problem otherwise but there is one point that I want to emphasize so which I will emphasize at the end in the in the discussion okay so what we have is that we have a simple rod a b cd this rod okay is supported by two sleeve supports okay though these are two sleeve supports which are small sleeves supported on these rigid rods and connected by pin to the main rod a b cd and on the top we have a string okay which is connected at point a and at the other point it is connected to a fixed support and the angle it makes with the horizontal alpha is equal to 30 degrees to this rod we are applying a force of 150 Newton and what we want to know is a tension in the wire and reaction at b and c now this is a sleeve support and if you remember what we had done is that if you just take the sleeve independently okay if the sleeve is completely free okay not connected to anything then this sleeve is free to slide like this because it is connected by pin to this rod okay the rotation of this rod with respect to the sleeve is also free and the sleeve is a small sleeve okay so there cannot be any reaction that the sleeve provides in this direction there cannot be any torque that the short sleeve can provide at this point only reaction that the sleeve can provide is that the sleeve is not is incapable of having any motion any possible motion in the direction perpendicular to the direction of these rods and so that will be the reaction that will act okay now we are asked to find out that for this system what is the tension in the wire and the reaction at b and c so there is a method which is much more quicker and there is a method which is not that difficult but which takes solving simultaneous equations okay so we will again get back to this problem in less than 10 minutes when I will briefly discuss this and we will move on to another problem okay so we are back okay a quick problem it is a very straight forward problem the free body diagram will look like this okay the free body diagram the free body diagram looks like this at point c okay just note what was this angle this angle was 30 degrees so the reaction will be perpendicular to this okay like this at point c this angle is 30 reaction will be perpendicular to this for the reason we had discussed okay this tension the tension will act along this direction and so this will be the free body diagram the simplest way to solve this diagram is this problem is to just assume a horizontal x axis and a vertical y axis and you can take moments about the points okay and some of all the forces but what will happen is that no matter what you do you will end up getting two simultaneous or three simultaneous equations you will never get an equation where there is only one variable that is present only one unknown variable present and p but instant what we do is that and this is a trick which we can keep playing again and again for certain problems what we ask ourselves is this what happens if the string were not present now if the string were not present what will happen is that this rod okay will try to slide in this direction okay so it will try to slide in this direction why because the two support reactions that are acting are in what direction perpendicular to this here and perpendicular to this okay so any force okay this p has a component which is which has any force which has a component okay this p has a component which is perpendicular to this okay these real support reactions won't be able to balance them so that is our trick to understand that what is appropriate coordinate system to use so what we do is that we define a coordinate system x which is parallel to this line 60 degrees and y which is perpendicular to this what we had just seen is that in the absence of the tension provided by this wire this rod has two reactions and this p component will have a component which is perpendicular to these two reactions which is p cos 60 and these will not balance them so then what we do is that we just take equilibrium along this inclined y direction and a component if you can see from geometry is that that t cos 60 will be balanced will balance the component of the force p cos 60 perpendicular to the wire and you balance them and you will immediately see that t is equal to p and once you get that then it's straight forward what you do is that for this free body diagram we already know what t is then take torque about point B you will immediately see that you will find what is the only fc t and p will not contribute only t and fc will contribute just do the moment balance you will see that fc is equal to minus 150 root 3 which means that the direction is opposite to what we had chosen similarly to find out what is the reaction at B we take torque about point C what do we get simple moment balance which we have discussed for ample number of times you will see that fB is equal to 300 root 3 which is acting in the direction that we have chosen so the simple point is that you can solve this problem by using x axis as the horizontal and y axis as a vertical but using geometric intuition what we realize is that that these two support reactions are parallel to each other and this applied force has a component that is perpendicular to these two parallel reactions and it is this tension which is balancing that which is balancing that particular contribution and so it is advantages for us to use a coordinate system which is inclined like this and advantage is that instead of getting simultaneous equations with more than one unknowns in any equation all our equations have exactly one unknown and because of exactly one another easier to solve and as a result the students will be less prone to making errors that is a only point of this question okay so we can take one quick question and then we'll move on to next problem which is a little bit more involved if you are considering a horizontal beam as for the sample problem that you have given there the loading and reactions you have the sample problem there sir okay the beam was horizontal and there was a he support there is a roller support also so at the hinge support hello at the hinge support we have two types of you have given that two components of the reactions only there but at that particular place since the beam is horizontal and all the loadings were particular yes so only one reaction is there to be there okay I see your point okay so this is a very common question which even students like for example you raised a very good point which is important to answer because if you take a beam like this there is a reason why we do that okay there is a reason it's a proper bookkeeping so we have a roller like this and you have a very valid point that since there is a roller and I don't have any horizontal force I only have a vertical force so when I draw the free body diagram why shouldn't I just draw this a y and b y why can't I just do that why should this be included why this is a and this is b and the reason is that it is simple bookkeeping if you have already the proper intuition you may as well say that this ax is equal to 0 but what happens is that then when you do that you should not use then the only two equations that are left to you then instead of 3 you are left with only two equations okay so if you can do that properly that when you remove this ax you are implicitly using the equation sigma fx equal to 0 and there is a chance that you may forget that afterwards and like the students may puzzle get puzzled and say oh that this is already 0 and we should have three equations but that third equation where has it gone so you have implicitly use that and for that confusion to not be there and to have appropriate bookkeeping what we say is that there is a possibility of ax and let me then use equilibrium in the x direction and then convince myself that that is equal to 0 rather than jumping the gun and immediately saying that it is equal to 0 because there may be some very subtle cases okay now this is horizontal rod there may some other subtle cases in which it may not be immediately clear that there should not be any horizontal reaction so it is just a decent practice to keep that ax and then later on say that my equilibrium equation in the horizontal direction makes it to be equal to 0 okay so we will now move on to the next problem okay so this was a very simple problem that was one point that I wanted to make the appropriate choice of coordinate axis can greatly simplify the problem that there is no hard and fast rule that we always should have x in the horizontal direction and y in the vertical direction by appropriately choosing the direction of x and y we can greatly simplify our problem and that was essentially the point of this question a simple question it was but this point is what we wanted to emphasize and what we want to tell the students that there is nothing sacred about horizontal and vertical for x and y choose what x and y is appropriate for that problem so that the problem become as simple as possible and as less prone to error as possible and in line with our physical intuition okay so we now move on to this problem okay it is a reasonably interesting problem okay it is a very interesting problem very simple but there is some amount of geometrical thinking that is involved in this problem so the only small trick in this problem okay so what we have here is we have a rod which is connected by two strings ac bc both real actual length is 4 meter 4 meter and length of ab is 4 meter ab car length is 6 meter now what we are doing is that we want to hoist this beam up by attaching a string pulling it over a pulley and applied a load p now hoisting means that when the beam is lying on the supports then there will be some reactions that are provided and when we want to hoist it this reactions will become 0 the only trick is that that when you lift this up what you realize is that that the simplest error that students do is that they assume that when the beam goes up ab remains horizontal now that is not possible for the simple reason is that that ac b when you lift it up will form a proper equilateral triangle and the line joining from c okay okay where the force p acts on it okay so this will have a tension of p and when this have a tension of p the line of action of p passes through a distance of 2 meters from support a but the weight of the beam acts at a distance of 3 meter from from this support a as a result this p and this w are not in the straight line and that will result a torque and that will end up inclining the rotating the beam like this so the idea is that that the beam will rotate by how much amount it is still an unknown that unknown is theta but how much will it rotate by it will rotate by an amount such that the load p acting at point c okay and this distance is always it will act in such a way that the center of gravity of the beam through which the load passes will be directly below this okay so that that is essentially the concept that the rotation should be such that this load peak a direction okay the line of action of load p should pass over the center of gravity of ac and then this is the final geometry of the problems the unknowns are ta tb and this angle theta now what do we do is this that these two are unknown so first we should know that this beam should be at equilibrium in the horizontal direction so ta cos this angle is 60- theta what a makes with the horizontal because this is theta so the angle at ta makes with the horizontal is 60- theta for b angle that tb makes with the horizontal is 60 plus theta and for equilibrium in the horizontal direction this is the condition ta cos 60- theta tb cos 60 plus theta now for this beam we can take moment balance about point B. So we will get one simple equation like this only involving ta and w second equilibrium equation will be Tb will have an equation like this. Now the unknowns present here are theta Tatb. So three equations three unknowns what we can do is that from these two equations will immediately see that Tb equal to 3ta and then use the first equation you will get an equation like this which you can very easily solve to find out that theta should be cos inverse of 12 by 13 and just substitute in one of these equations you will see that ta equal to w by root 13 and Tb is equal to 3w by root 13. So the only trick present in this problem is that that AB cannot remain horizontal because if AB remains horizontal the line of action of P does not pass through the line of action of w and as a result there will be a net torque on the beam and the beam cannot stay in equilibrium. So it has to come to equilibrium in some inclined position like this and inclination of the beam is also another problem in this beam. So three equation three unknowns and our consistency mathematical consistency is taken care of here. So if you have any queries we will have some discussions for 2, 3 more minutes before moving on to the next problem. 1, 1, 9, 2 go ahead we can hear you, I have a question on equilibrium of the body sir. Okay please. As we have seen when the different forces as we have seen when the different forces acting on the body nullify each other the body remains in equilibrium. Okay that is only one part because all the forces and the moments both should like be nullified some of all the moments acting on a free body and a sum of all the moments external and moments created by force both should nullify only and only then the body is in equilibrium not just the forces also the moments. Now yeah yeah yes forces and moments when they are nullified the body remains in equilibrium now my question is yes when earth is a rotating body now it is rotating about its own axis now how we will explain the equilibrium of such type of body like earth. Oh yeah so those you know what these are the problems which we should not worry about with this these are the points about they are called as moving reference frame systems and typically strictly speaking okay so there is like our earth is a non-inertial frame of reference so strictly speaking we should not be talking about all this equilibrium business and so on but what we assume is that for our daily purposes okay that the rotation speed is so small okay that like for example for all practical purposes okay for all our equilibrium on our earth those Coriolis forces and the other forces okay so that is beyond the scope of this course okay that we neglect those forces okay totally we neglect those forces but if you really write down equations of equilibrium properly all those forces that come from the rotation of the earth should also be taken into account okay and we will see that we are strictly speaking not in equilibrium because we are rotating and any rotation would involve like lot of forces centripetal forces Coriolis forces and what not okay so that is beyond the scope of this course here we are assuming that we are in some kind of a local shell in which the rotation effects for example and they are not that dominant they are important but for all the practical purposes the all the other forces that are involved this particular force is much smaller compared to the other forces that are involved so we neglect it but you have a very valid point but that is beyond the scope of this course okay so earth is a non-inertial frame of reference okay we are already in deep waters there 1 2 5 3 the doubt is that in case we have 4 unknowns okay and 3 and 3 equations how to solve analytically such kind of problems we pray to God okay because 4 equations 3 unknowns okay you cannot solve it and these are called as examples of statically indeterminate systems I am not saying that you cannot do anything about it for example to in tomorrow's tutorial you will see that there are certain statically indeterminate problems where the number of equations is actually less than the number of unknowns but the geometry of the problem is such that you can obtain certain forces but clearly when the number of equations okay is less than the number of unknowns you cannot obtain all the unknowns okay that is the general theorem in linear algebra so either you have no solution or you may have multiple solution okay so it depends on what the equations are but typically speaking you will not have a unique solution and these kind of problems typically come under the purview of structural mechanics strength of material where you use what is what are called as compatibility conditions and then you attempt and get extra equations from displacement criteria using principle of virtual work and so on and then solve those problems but as far as e make is concerned we are not using those techniques so within e make if you get four unknowns and three equations by enlarge okay the train stops there okay so we are done with this problem so the one question which was asked about earth being rotating and a non-inertial term is a very interesting question only thing is that the forces that come from those non-inertial rotation of the earth and so on they are typically much smaller compared to the weight of the body or other forces that are involved and so we do not need to bother with them as far as our general engineering problems are concerned but there are certain applications which are so sensitive for which you need to take into account all those forces and a kind of problems and a kind of values you need to input or the kind of the sensitivity you need in that analysis is far more but whereas in our course we are looking at a very simple set of problems where the problem themselves may be kind of non-trivial but all these subtle issues coming from the rotation of the earth or our rotation around the sun those things okay are much smaller compared to the other forces involved and the kind of sensitivity we want in our calculations so moving on to the next problem okay so there are many other problems so three four more problems are there okay so you can solve them at your leisure okay when you go back but there is one simple problem that we are trying to understand is a uniform 400 kg drum so you have a uniform drum okay a uniform 400 kg drum is mounted on a line of rollers so we have a line of rollers okay so the drum is like a cylinder and there is a line of rollers what is the line so this is the plane through which I am looking so the line of rollers is coming out of plane so there is a line of rollers at A and there is a line of rollers at B now what is asked is that 80 kg man moves slowly a distance of 700 millimeters okay the geometry is given to you that the man's center of gravity lies at a distance of 700 millimeters from point O and all the rollers are perfectly free to rotate now this is the crux of the problem that at A all the rollers are perfectly free to rotate what does that mean if they are all perfectly free to rotate then they cannot have any lateral friction they can only exert a normal load at point A whereas at B all the rollers are free to rotate means all the rollers at B are also free to rotate which means that they can exert only a normal reaction except one okay which must overcome appreciable friction in its bearing what is a bearing that the roller is it is like a pulley and it is connected to the support by some bearing and there is lot of friction between that roller and the bearing what does that mean that the force that that one particular roller that will exert on this drum will have one horizontal component and one vertical component okay and which must overcome appreciable friction in its bearing so it has capacity of producing because it is not free to rotate now it has a capacity of producing both a horizontal force as well as a vertical force now with this in mind what we are asked is that calculate the friction force exerted by that one roller tangent to the drum okay that what is that tangential force that is exerting on this drum as well as the magnitude R okay of exerted by all the rollers at A on this drum for this condition okay so A all the rollers can exert only a normal reaction here B all the rollers can exert except one which can also exert a tangential force in addition to normal force so we want to know what is the total normal force by all the rollers on the drum and what is the friction force that that one stuck roller is exerting on this on this drum at point B given this is the configuration of the person and the drum okay so I think many many of us got the answer it is a reasonably straight forward problem but the reason I decided to discuss this problem that even though it is not too difficult the only thing is that that we have to ensure that there are two free bodies here okay so strictly speaking it is not a one body problem that this person okay inside that is another free body diagram so what we want to know is what is the reaction that this person exerts okay at this point now note that if you draw the free body diagram for that person there may be whatever reactions coming from the contact of this person with the cylinder but weight acts downward 80 kgs and for the equilibrium of this person itself no matter what it does okay the total reaction can only be vertical and in line with this 80 Newton as a result is 700 meter where the center of gravity of the person is present okay should itself be the line at which the effective reaction that the person exerts on the cylinder because we draw the free body diagram and realize that whatever is the weight of the person equal and opposite and along the same line should be the reaction of the cylinder on this person and then by Newton's third law this 80 should act on the cylinder and this is our free body diagram for the cylinder and then afterwards it is straight forward that what is going to happen that if you take this cylinder these normal reactions pass about the center now the weight of the cylinder is supposed to be the cylinder which is supposed to be homogeneous also passes through point O and as a result now what happens is that that this force F okay is the only way in which the torque that the reaction created by the 80 kgs of the person can be balanced if there were no F then all these three other forces pass through O and the torque that the person exerts on the cylinder there is no way to balance that so it is that stuck roller here which is providing this friction F that is balancing this and in straight forward we take torque for this free body diagram about point O we get that W into 700-FE to 800 should 1800 equal to 0 we get immediately F and to find out what is the reaction of all the set of rollers at point A we can do one thing we can take for this free body diagram what are some of all the torques of all the loads about point B and then we can immediately obtain what is RA or another thing note that this is 60 is 30 so RA is perpendicular to RB so we already found out F what we can do is that we can take equilibrium force equilibrium in the inclined direction and what you will see that you do whatever way you will get the same answer and that is a cross check that you can have at your disposal problem 5 is also a little bit of a challenging problem okay so let me quickly discuss this problem okay before we retire for the day it is a little bit different and whatever we have discussed till now so what we have is that again we have a semi circular rod okay ABCD which is supported by frictionless rollers okay at point B and at point C and there is another roller support at point D so what does this mean is that this support B can exert a reaction only in a direction perpendicular but only outwards so the direction of the reaction exerted B on this rod can only be towards O and the direction of the reaction that is exerted at point C can only be towards O it cannot exert a reaction which is in the away from O because why because that is not a contact here it is for example it is situated it is placed on this roller so it can only press against it but cannot lose contact with it and now what we are asked is that that for this system you are applying a load P at some angle theta and you are asked to find out what is the maximum theta that is allowed at this point okay such that this system remains in equilibrium okay so that is the point okay but what we can do is that before that we can have a two-minute discussion session on the previous problem any question from 1 2 3 5 we can here you go ahead sir my doubt is about joint sir okay we join sir in a in a pasted joint how can you take this sir for example cricket bat with cricket bat with the handle the handle is pasted with the bat sir right sir yes so how can we consider that joint sir is it a hinge it or it has to be a fixed joint because if it is not a fixed joint it is a hinge joint then the handle can rotate about the bat right the main body of the bat so it has to be a fixed joint if you want to just model it like a simple e-mail problem there will be distributed forces acting at the contact point so if you look at if you draw a free world diagram for a handle this is the main part of the meat of the bat and this is the handle so if you draw the free body diagram only for the handle okay and there are forces acting on this handle then in that case there will be a lot of distributed forces friction and other things between the handle and the main part of the bat but effectively if you look at statically equivalent system this will look like this okay there will be lot of distributed forces but the effect of that as far as the dynamics of rigid body is concerned or equilibrium of rigid body is concerned is this will be equivalent to three forces and a torque like this okay but I am having another thought sir so is there any difference between normal bearing and lengthy bearing sir so I have met up met with the problem actually by considering the normal bearing we are taking that as that as a simply supported okay which by considering lengthy bearing we are taking that as a fixed support is it right sir so I can understand what but it you are right I agree with you okay so for example you can have a sleeve like this a big sleeve okay and you can have a beam like this right you are saying like this right and this is fixed like this so you are totally right if for example this leaf is relatively large then what happens is that that you are restricting the rotation of this beam as far as this joint is concerned and if you draw the free body diagram of this what will happen is that this point will connect here this point will connect here and if this sleeve is reasonably large then these two forces can create a torque and the equivalent system then has to be written like this this reaction may not be there depending on if there is a friction or not so it will be something like this so you are totally right if the sleeve is large then what can happen is that that this point of the beam can contact here this point can contact here and that combination may lead to a equivalent torque and so when we draw the free body diagram we may have to replace that with a torque like this I agree with you thank you sir yeah thank you so in this problem the concept is as simple as possible is as simple as this that at support B the reaction can only act inwards at support C the reaction can act only inwards okay so what we decide is that that our load acting at an angle can be broken into two parts P y and P x so let us analyze these two problems separately to figure out what may be the issue that can creep in that why do we need restriction on the angle what we see here is that that what we emphasize the students is that since the reaction can only act inwards so let us look at this problem so you can immediately figure out by taking moment about point O that F d y should be 0 here now this F B will be a combination of F B 1 and F B 2 where F B 1 comes from applying a load horizontally and F B 2 comes from applying load vertically same for the reaction at C now you note here is that that when a horizontal load is applied this reaction can immediately be figured out to be 0 by taking torque about point O and then what you will figure out by this by taking torque about point C that F B 1 is equal to minus P x so F B 1 is not inverse okay but it has to be outwards but that is not possible this support B will not allow you to exert a reaction on this in the in the outward direction it can only exert a reaction like this because otherwise when this is negative this rod will lose contact with the support but now what happens is that if you apply a load P in the vertical direction P y then you will see that the reaction at this point okay F B 2 is positive and given by 2 root P y by 3 whereas here it is negative and given by minus P x you can convince we can convince ourselves that the reaction at C is always positive so what is happening is that this load in the horizontal direction is trying to make this rod disrupt equilibrium about point B where whereas the vertical load is only stabilizing and the combination is F B is equal to F B 1 plus F B 2 which now has to be positive for this thing to be in equilibrium because if this becomes negative it means that the rod is going to lose contact at support B and for this to be greater than 0 this is the condition that we get P x can be written as P sin theta P y can be written as P cos theta and this greater than equal to 0 will give us a condition that tan theta should be lose than 2 by root 3 which is approximately 49 degrees so that is the crux of the problem that you apply horizontal load this support tends to get disrupted vertical load the support is stabilized when you apply combination the combination can be such that the total force at F B should be positive inverse and if you increase the angle you will see that this becomes dominant and the support and the rod will lose equilibrium at contact B